If $y=\cot^{-1}\!\left[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$, find $\dfrac{dy}{dx}$.

CBSE Class 12 · 2008 · DifferentiationLong Answer

y = cot^{- 1} [\frac{1 + sin x + 1 - sin x}{1 + sin x - 1 - sin x}]

, find

\frac{d y}{d x}

Answer: $\frac{1}{2}$

Write

1 \pm sin x = (cos \frac{x}{2} \pm sin \frac{x}{2})^{2}

. For

0 < x < \frac{π}{2}

1 + sin x = cos \frac{x}{2} + sin \frac{x}{2}, 1 - sin x = cos \frac{x}{2} - sin \frac{x}{2}

Then the bracket becomes:

\frac{2 cos \frac{x}{2}}{2 sin \frac{x}{2}} = cot \frac{x}{2}

Therefore:

y = cot^{- 1} (cot \frac{x}{2}) = \frac{x}{2}

Differentiate:

\frac{d y}{d x} = \frac{1}{2}

CBSE Class 12 Board · 2008 · Subjective